Statistics on the number of closed Islands

This problem didn't work out , I read the solution for a long time , First of all, there are not two ferrules , And four 1 not .

I really didn't think out how to do it . What is closed , It means that there is no 1. With this sentence came out . In this case , Just go through all 0, towards 0 Up, down, left and right traversal of , If you go to the edge, it is not considered closed . The nodes traversed at the same time should be marked , Prevent multiple iterations , Here is the solution to the problem of learning 1, If I want to, it must be 2, Avoid repetition .

#include<stdc++.h>

using namespace std;
bool is_island;

void dfs(vector<vector<int>>&grid, int x, int y)
{
	if(x<0 || x>=(int)grid.size() || y<0 || y>=(int)grid[0].size())
	{
		is_island = false;
		return;
	}

	if(grid[x][y]==0)
	{
		grid[x][y] = 1;
		dfs(grid, x+1, y);
		dfs(grid, x, y+1);
		dfs(grid, x-1, y);
		dfs(grid, x, y-1);
	}
}

int closedIsland(vector<vector<int>>& grid) {
	int count = 0;
	for(int i=1; i<(int)grid.size()-1;i++)
	{
		for(int j=1; j<(int)grid[0].size()-1; j++)
		{
			if(grid[i][j]==0)
			{
				is_island = true;
				dfs(grid, i,j);
				if(is_island)
					count++;
			}
		}
	}
	return count;
}
int main()
{
	int x[5][8] = {
   {1,1,1,1,1,1,1,0},{1,0,0,0,0,1,1,0},{1,0,1,0,1,1,1,0},{1,0,0,0,0,1,0,1},{1,1,1,1,1,1,1,0}};
	vector<vector<int>> a;
	for(int i=0; i<5;i++)
	{
		vector<int>b;
		for(int j=0; j<8;j++)
		{
			b.push_back(x[i][j]);
		}
		a.push_back(b);
	}
	cout<<closedIsland(a)<<endl;
	return 0;
}